Complete the hexagonal and star-shaped Rangolies [see Fig. $(i)$ and $(ii)$] by filling them with as many equilateral triangles of side $1\,cm$ as you can. Count the number of triangles in each case. Which has more triangles?
(B) To find the number of equilateral triangles of side $1\,cm$ that can fit into the given shapes,we calculate the area of the shapes and divide by the area of one small equilateral triangle.
$1$. For the regular hexagon of side $s = 5\,cm$:
Area $= \frac{3\sqrt{3}}{2} s^2 = \frac{3\sqrt{3}}{2} \times 25 = 37.5\sqrt{3} \approx 64.95\,cm^2$.
Each small triangle of side $1\,cm$ has an area of $\frac{\sqrt{3}}{4} \times 1^2 = 0.25\sqrt{3} \approx 0.433\,cm^2$.
Number of triangles in Fig. $(i) = \frac{37.5\sqrt{3}}{0.25\sqrt{3}} = 150$.
$2$. For the star-shaped figure in Fig. $(ii)$:
This figure consists of a central hexagon of side $5\,cm$ and $6$ equilateral triangles of side $5\,cm$ attached to its sides.
Total area $= \text{Area of hexagon} + 6 \times \text{Area of triangle of side } 5\,cm$.
Total area $= 37.5\sqrt{3} + 6 \times (6.25\sqrt{3}) = 37.5\sqrt{3} + 37.5\sqrt{3} = 75\sqrt{3} \approx 129.9\,cm^2$.
Number of triangles in Fig. $(ii) = \frac{75\sqrt{3}}{0.25\sqrt{3}} = 300$.
Therefore,the star-shaped figure $(ii)$ has more triangles.
$1$. For the regular hexagon of side $s = 5\,cm$:
Area $= \frac{3\sqrt{3}}{2} s^2 = \frac{3\sqrt{3}}{2} \times 25 = 37.5\sqrt{3} \approx 64.95\,cm^2$.
Each small triangle of side $1\,cm$ has an area of $\frac{\sqrt{3}}{4} \times 1^2 = 0.25\sqrt{3} \approx 0.433\,cm^2$.
Number of triangles in Fig. $(i) = \frac{37.5\sqrt{3}}{0.25\sqrt{3}} = 150$.
$2$. For the star-shaped figure in Fig. $(ii)$:
This figure consists of a central hexagon of side $5\,cm$ and $6$ equilateral triangles of side $5\,cm$ attached to its sides.
Total area $= \text{Area of hexagon} + 6 \times \text{Area of triangle of side } 5\,cm$.
Total area $= 37.5\sqrt{3} + 6 \times (6.25\sqrt{3}) = 37.5\sqrt{3} + 37.5\sqrt{3} = 75\sqrt{3} \approx 129.9\,cm^2$.
Number of triangles in Fig. $(ii) = \frac{75\sqrt{3}}{0.25\sqrt{3}} = 300$.
Therefore,the star-shaped figure $(ii)$ has more triangles.
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